Question: Evaluate the double integral. $ \int_0^\pi \int_0^{\sin(y)} 2x - \sin(y) \, dx \, dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $\pi^2$ (Choice C) C $\pi - 2\pi^2$ (Choice D) D $0$
Solution: First, we evaluate the inner integral. We can substitute in the $\sin(y)$ at the end as if it were a numerical bound. $\begin{aligned} & \int_0^\pi \int_0^{\sin(y)} 2x - \sin(y) \, dx \, dy \\ \\ &= \int_0^\pi \left[ x^2 - x\sin(y) \right]_0^{\sin(y)} \, dy \\ \\ &= \int_0^\pi \sin^2(y) - \sin^2(y) \, dy \\ \\ &= \int_0^\pi 0 \, dy \end{aligned}$ Second, we evaluate the outer integral, which is zero. $\begin{aligned} \int_0^\pi 0 \, dy = 0 \end{aligned}$ The answer: $ \int_0^\pi \int_0^{\sin(y)} 2x - \sin(y) \, dx \, dy = 0$